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3 years ago

A body of mass m1 = 1.5 kg moving along a directed axis in the positive sense with a velocity

Physics

1 answer:

larisa86 [58]3 years ago

8 0

Answer:

3.71 m/s in the negative direction

Explanation:

From collisions in momentum, we can establish the formula required here which is;

m1•u1 + m2•v2 = m1•v1 + m2•v2

Now, we are given;

m1 = 1.5 kg

m2 = 14 kg

u1 = 11 m/s

v1 = -1 m/s (negative due to the negative direction it is approaching)

u2 = -5 m/s (negative due to the negative direction it is moving)

Thus;

(1.5 × 11) + (14 × -5) = (1.5 × -1) + (14 × v2)

This gives;

16.5 - 70 = -1.5 + 14v2

Rearranging, we have;

16.5 + 1.5 - 70 = 14v2

-52 = 14v2

v2 = - 52/14

v2 = 3.71 m/s in the negative direction

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2 years ago

Find the density of seawater at a depth where the pressure is 500 atm if the density at the surface is 1100 kg/m^3 . Seawater ha

mixer [17]

The density of seawater at a depth where the pressure is 500 atm is $1124kg/m^3$

Explanation:

The relationship between bulk modulus and pressure is the following:

$B=\rho_0 \frac{\Delta p}{\Delta \rho}$

where

B is the bulk modulus

$\rho_0$ is the density at surface

$\Delta p$ is the variation of pressure

$\Delta \rho$ is the variation of density

In this problem, we have:

$B=2.3\cdot 10^9 N/m^2$ is the bulk modulus

$\rho_0 =1100 kg/m^3$

$\Delta p = p-p_0 = 500 atm - 1 atm = 499 atm = 5.05\cdot 10^7 Pa$ is the change in pressure with respect to the surface (the pressure at the surface is 1 atm)

Therefore, we can find the density of the water where the pressure is 500 atm as follows:

$\rho = \rho_0 + \Delta \rho = \rho_0+\frac{\rho_0 \Delta p}{B}=\rho_0 (1+\frac{\Delta p}{B})=(1100)(1+\frac{5.05\cdot 10^7}{2.3\cdot 10^9})=1124kg/m^3$

Learn more about pressure in a fluid:

brainly.com/question/9805263

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3 years ago

A motorcyclist changes the velocity of his bike from 20.0 meters/second to 35.0 meters/second under a constant acceleration of 4

Fiesta28 [93]

35-20 = 15m/s difference
15/4 = 3.75 seconds

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3 years ago

A light ray moving at a 54.3 deg

MrMuchimi

Answer:

$45.25^{\circ}$

Explanation:

Applying Snell's law

$n_1sin\theta_1 = n2*sin\theta_2\\1.33*sin(54.3) = 1.52*sin\theta_2\\sin\theta_2 = 1.33*0.81208353/1.52 = 0.71057\\\theta_2=sin^{-1} (0.71057)\\\theta_2 = 45.25^{\circ}$

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3 years ago

Which of the following statements about the problem of object recognition is false?

alexandr1967 [171]

Answer:

The answer is: letter c, in object recognition, the goal is recognizing the proximal stimulus.

Explanation:

Letter c is a "false" statement about object recognition because the goal is recognizing the distal stimulus and "not the proximal stimulus."

Distal stimulus refers to <em>an event or an object in the world that provides information to the proximal stimulus. </em>The proximal stimulus is a pattern of these events and objects that reaches to your senses. They can be registered in the person via<em> "sensory receptors." </em>

We need to recognize the distal stimulus and not the proximal stimulus. For example, when a lemon (distal stimulus) is being cut, it brings out a fragrance (proximal stimulus) that goes to the person's sense of smell. This gives the person a hint on where the smell is coming from and what it is. Then, the person recognizes that it is a lemon.

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3 years ago